I came up with this theorem (shown below) a few months ago, and I haven't been able to find anything like it on the web. This theorem will give you the quadratic expression that results from $(x-(a+bi))(x-(a-bi))$. This theorem is useful for finding all roots of an expression when given a complex root (because complex roots come in conjugate pairs, remember?). Can anyone tell me if this is old news, or if I have discovered something?
Some context:
I'm the student not the teacher. My teacher (which, BTW, she's my Mom, since I'm homeschooled) is trying to get me to re-prove every time.
Basic Theorem (which if it doesn't have a name, I'll call it Smith's Product of Complex Conjugate Roots Theorem. A tad long, don't you think?):
$(x-(a+bi))(x-(a-bi)) = x^2-2ax+a^2+b^2$
Proof 1:
$(x-(a+bi))(x-(a-bi))$
$(x-a-bi)(x-a+bi)$
$x^2-ax+bxi-ax+a^2-abi-bxi+abi-b^2i^2$
$x^2-2ax+a^2-b^2i^2$
$x^2-2ax+a^2+b^2$
Proof 2:
$(x-(a+bi))(x-(a-bi))$
$(x-a-bi)(x-a+bi)$
$((x-a)-bi)((x-a)+bi)$
$(x-a)^2-b^2i^2$
$(x^2-2ax+a^2)+b^2$
$x^2-2ax+a^2+b^2$